leetcode 第132场周赛

https://leetcode.com/contest/weekly-contest-132

对应题目：1025-1028

 

1：博弈问题，后继节点存在一个为败，当前为即为必胜。

    public boolean divisorGame(int N) {
        boolean[] dp = new boolean[1005];
        dp[1] = false;
        dp[2] = true;
        dp[3] = false;
        for (int i = 4; i <= N; i++) {
            boolean f = false;
            for (int j = 1; j < i; j++) {
                if(i % j == 0 && !dp[i - j]) {
                    f = true;
                    break;
                }
            }
            dp[i] = f;
        }
        return dp[N];
    }

2：dfs，维护最大值和最小值

    	int ret = 0;
	    public int maxAncestorDiff(TreeNode root) {
	    	ret = 0;
	        dfs(root, root.val, root.val);
	        return ret;
	    }
	    
	    void dfs(TreeNode cur, int max, int min)
	    {
	    	if(cur == null)return;
	    	ret = Math.max(ret, Math.abs(max-cur.val));
	    	ret = Math.max(ret, Math.abs(min-cur.val));
	    	
	    	dfs(cur.left, Math.max(max, cur.val), Math.min(min, cur.val));
	    	dfs(cur.right, Math.max(max, cur.val), Math.min(min, cur.val));
	    }

 

3：找到子数组（可以不连续）长度最大的等差数列

dp[i][d]表示以i结尾的公差为d的最大长度: dp[i][d] = dp[j][d] + 1(j < i)

d可能为负数，加个值修正一下。

    public int longestArithSeqLength(int[] A) {
        int ans = 2;
        int n = A.length;
        int diff = 10000;
        int[][] dp = new int[10010][20010];
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                int d = A[i] - A[j];
                int ddiff = d + diff;
                if (dp[j][ddiff] == 0) {
                    dp[i][ddiff] = 2;
                } else {
                    dp[i][ddiff] = dp[j][ddiff] +1;
                }
                ans = Math.max(ans, dp[i][ddiff]);
            }
        }
        return ans;
    }

4：给了一个树的深度遍历，还原树。

按照题意遍历，贴个uwi的代码

public class D {
    int pos = 0;
    char[] s;
    int len = 0;

    public TreeNode recoverFromPreorder(String S) {
        len = S.length();
        s = S.toCharArray();
        pos = 0;
        return go(0);
    }

    TreeNode go(int dep) {
        int v = 0;
        while (pos < len && s[pos] >= '0' && s[pos] <= '9') {
            v = v * 10 + (s[pos] - '0');
            pos++;
        }
        TreeNode cur = new TreeNode(v);

        if (hasEdge(dep + 1)) {
            pos += dep + 1;
            cur.left = go(dep + 1);
        }
        if (hasEdge(dep + 1)) {
            pos += dep + 1;
            cur.right = go(dep + 1);
        }

        return cur;
    }

    boolean hasEdge(int d) {
        if (pos + d > len - 1) return false;
        for (int i = pos; i < pos + d; i++) {
            if (s[i] != '-') return false;
        }
        return s[pos + d] != '-';
    }
}